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I'm embarrassed to ask this question because I feel that I should know the answer.

What I can't figure out how to do consistently is calculate feed point impedances.

Particularly something as simple as a dipole with a tuned line. How do you all calculate the impedance at the end of the feed?

Are you using EZNEC? I have the demo and I can't figure out how to do this.

How would I do it manually? That's what would really teach me.

Is there a step by step process someone could walk me through?

If I just need to read page xxx-xyz of the handbook you can say that too... But please tell me exactly what I need to look at!

2. Ham Member
Join Date
Mar 2007
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Charley,

You can do it in EZNEC by connecting a Transmission Line element to the dipole feedpoint, entering the TL details - length, Zo, loss, etc - and then moving the Source to the far end of the TL. If you have the latest version of EZNEC you terminate the TL on a "virtual wire" and place the Source there.

However, it's probably easier - and more accurate - to get the antenna feedpoint impedance using EZNEC, and then plug that impedance into VK1OD's transmission line calculator:
http://vk1od.net/calc/tl/tllc.php

The most instructive way is probably to get the dipole feedpoint impedance using EZNEC, plot that on a Smith chart, and then "rotate" around the chart the appropriate amount for your feedline length.

Good luck.

Steve G3TXQ

4. Originally Posted by AE2CS
I'm embarrassed to ask this question because I feel that I should know the answer. ...
Don't say that. Your questions do not fall into the "Who was buried in Grant's tomb?" category. :-)

5. Ham Member
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Mar 2007
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Charley,

If you really want to do the calculation for yourself, here's the simplified formula which applies to lossless lines:

Z = Z0(ZL + jZ0 tan(kl)) /(Z0 + jZL tan(kl))

Z is the input impedance
Z0 is the characteristic impedance of the line
kl is the length of the line in electrical degrees (e.g quarter-wave = 90 degrees)

Remember all the Zs are complex quantities.

Steve G3TXQ

6. Ham Member
Join Date
Nov 2002
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1,049
Charley,
There is a good reason for this misunderstanding. There is an ARRL article titled, "My Feed line Tunes My Antenna" The first example asks about a 50 Ohm resistor at the end of a piece of 50 ohm surge impedance coax. Indeed the impedance is 50 ohms at the other end. However, Once the load changes to something other than 50 ohms resistive the best word to describe what happens is "Complex"! Read the article, it will be a good start.
73,
Rob WA9UAA

7. Originally Posted by G3TXQ
Charley,

If you really want to do the calculation for yourself, here's the simplified formula which applies to lossless lines:

Z = Z0(ZL + jZ0 tan(kl)) /(Z0 + jZL tan(kl))

Z is the input impedance
Z0 is the characteristic impedance of the line
kl is the length of the line in electrical degrees (e.g quarter-wave = 90 degrees)

Remember all the Zs are complex quantities.

Steve G3TXQ
This, and the other responses, are excellent. Thank you. I will be working on this.

8. Ham Member
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Mar 2007
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Charley,

Here are some practical numbers if you want to try the "long hand" calculation.

EZNEC says the feedpoint impedance of a G5RV used on 80m is about 30-j350

The ladderline is 300 Ohms characteristic impedance, and is 47.5 electrical degrees long on 80m.

Tan(47.5) = 1.09, so your starting point is:

Zin = 300 x [(30-j350) + j300 x 1.09] / [300 + j(30-j350) x 1.09]

You should end up somewhere near: Zin = 12.7 - j10.7

The "sanity check" is that impedance is a VSWR(50) of just over 4:1, which looks right for a G5RV used on 80m.

73,
Steve G3TXQ

9. With an equation like this - does it make more sense to convert to polar from the rectangular coordinates?

Its been a while since I had to do any vector math or more complicated algebra! I'm probably making it harder than it is. We have a complex number multiplied by something that is not a complex number.

I'm flying 300kph and heading 45 degrees - if someone says "multiply that quantity" by 1.09 do I keep heading 45 but increase my speed to 327kph?

I know if it were "multiply by 1.09 | 10 " then I would multiply my speed and add my bearing (300 | 45) (1.09 | 10) = 327 | 55

Should I convert the rectangular to polar and then treat the 1.09 as 1.09 | 0?

10. Ham Member
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Mar 2007
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Originally Posted by AE2CS
With an equation like this - does it make more sense to convert to polar from the rectangular coordinates?
Charley,

It's easier to stay in Rectangular coordinates when you are adding and subtracting complex numbers, and easier to drop into Polar form when doing multiplication or division.

To answer your question: if you multiply a Polar form complex number by a scaling factor, you scale the vector length but don't change the angle.

It might make things clearer if I work through that example:

Zin = 300 x [(30-j350) + j300 x 1.09] / [300 + j(30-j350) x 1.09]

Zin = 300 x [30 - j350 + j327] / [300 +j32.7 + 381.5]

Zin = 300 x [30 - j23] / [681.5 + j32.7]

Here's the point you might want to swap into Polar form using:

[30 - j23] = 37.8 Angle -37.5
[681.5 + j32.7] = 682.3 Angle 2.7

Then:

Zin = 300 x 37.8<-37.5 / 682.3<2.7

Zin = 16.6 Angle -40.2

Converting back to Rectangular form

Zin = 12.7 -j10.7

Hope that helps.

Many modern scientific calculators have complex number capability, so that makes it a lot easier. And of course it's even easier to let EZNEC do it all for you

Steve G3TXQ

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